Due by 11:59pm on Wednesday, 11/4

## Instructions

Download hw08.zip. Inside the archive, you will find a file called hw08.scm, along with a copy of the OK autograder.

Submission: When you are done, submit with ```python3 ok --submit```. You may submit more than once before the deadline; only the final submission will be scored. See Lab 1 for instructions on submitting assignments.

Using OK: If you have any questions about using OK, please refer to this guide.

Readings: You might find the following references useful:

### Question 1

Write a procedure `substitute` that takes three arguments: a list `s`, an `old` word, and a `new` word. It returns a list with the elements of `s`, but with every occurrence of `old` replaced by `new`, even within sub-lists.

``````(define (substitute s old new)
'YOUR-CODE-HERE
)``````

Use OK to unlock and test your code:

``````python3 ok -q substitute -u
python3 ok -q substitute``````

### Question 2

Write `sub-all`, which takes a list `s`, a list of `old` words, and a list of `new` words; the last two lists must be the same length. It returns a list with the elements of `s`, but with each word that occurs in the second argument replaced by the corresponding word of the third argument.

``````(define (sub-all s olds news)
'YOUR-CODE-HERE
)``````

Use OK to unlock and test your code:

``````python3 ok -q sub-all -u
python3 ok -q sub-all``````

### Differentiation

The following problems develop a system for symbolic differentiation of algebraic expressions. The `derive` Scheme procedure takes an algebraic expression and a variable and returns the derivative of the expression with respect to the variable. Symbolic differentiation is of special historical significance in Lisp. It was one of the motivating examples behind the development of the language. Differentiating is a recursive process that applies different rules to different kinds of expressions:

``````; Derive returns the derivative of exp with respect to var.
(define (derive expr var)
(cond ((number? expr) 0)
((variable? expr) (if (same-variable? expr var) 1 0))
((sum? expr) (derive-sum expr var))
((product? expr) (derive-product expr var))
((exp? expr) (derive-exp expr var))
(else 'Error)))``````

To implement the system, we will use the following data abstraction. Sums and products are lists, and they are simplified on construction:

``````; Variables are represented as symbols
(define (variable? x) (symbol? x))
(define (same-variable? v1 v2)
(and (variable? v1) (variable? v2) (eq? v1 v2)))

; Numbers are compared with =
(define (=number? expr num)
(and (number? expr) (= expr num)))

(define (make-sum a1 a2)
(cond ((=number? a1 0) a2)
((=number? a2 0) a1)
((and (number? a1) (number? a2)) (+ a1 a2))
(else (list '+ a1 a2))))
(define (sum? x)
(and (list? x) (eq? (car x) '+)))

(define (make-product m1 m2)
(cond ((or (=number? m1 0) (=number? m2 0)) 0)
((=number? m1 1) m2)
((=number? m2 1) m1)
((and (number? m1) (number? m2)) (* m1 m2))
(else (list '* m1 m2))))
(define (product? x)
(and (list? x) (eq? (car x) '*)))

### Question 3

Implement `derive-sum`, a procedure that differentiates a sum by summing the derivatives of the `addend` and `augend`. Use data abstraction for a sum:

``````(define (derive-sum expr var)
'YOUR-CODE-HERE
)``````

Use OK to unlock and test your code:

``````python3 ok -q derive-sum -u
python3 ok -q derive-sum``````

### Question 4

Implement `derive-product`, which applies the product rule to differentiate products:

``````(define (derive-product expr var)
'YOUR-CODE-HERE
)``````

Use OK to unlock and test your code:

``````python3 ok -q derive-product -u
python3 ok -q derive-product``````

### Question 5

Implement a data abstraction for exponentiation: a `base` raised to the power of an `exponent`. The `base` can be any expression, but assume that the `exponent` is a non-negative integer. You can simplify the cases when `exponent` is `0` or `1`, or when `base` is a number, by returning numbers from the constructor `make-exp`. In other cases, you can represent the exp as a triple `(^ base exponent)`.

Hint: The built-in procedure `expt` takes two number arguments and raises the first to the power of the second.

``````; Exponentiations are represented as lists that start with ^.
(define (make-exp base exponent)
'YOUR-CODE-HERE
)

(define (base exp)
'YOUR-CODE-HERE
)

(define (exponent exp)
'YOUR-CODE-HERE
)

(define (exp? exp)
'YOUR-CODE-HERE
)

(define x^2 (make-exp 'x 2))
(define x^3 (make-exp 'x 3))``````

Use OK to unlock and test your code:

``````python3 ok -q make-exp -u
python3 ok -q make-exp``````

### Question 6

Implement `derive-exp`, which uses the power rule to derive exps:

``````(define (derive-exp exp var)
'YOUR-CODE-HERE
)``````

Use OK to test your code:

``python3 ok -q derive-exp``